Question

If the diagonals of the quadrilateral formed by the lines l₁x + m₁y + n₁ = 0, l₂x + m₂y + n₂ = 0, l₁x + m₁y + n₁' = 0 and l₂x + m₂y + n₂' = 0 are perpendicular, then write the value of l₁² − l₂² + m₁² − m₂².

Answer 1

The given lines are
l₁x + m₁y + n₁ = 0      ... (1)
l₂x + m₂y + n₂ = 0      ... (2)
l₁x + m₁y + n₁' = 0     ... (3)
l₂x + m₂y + n₂' = 0     ... (4)
Let (1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

The equation of diagonal AC passing through the intersection of (2) and (3) is given by
l₁x + m₁y + n₁' + \[\lambda\] (l₂x + m₂y + n₂) = 0

\[\Rightarrow \left( l_1 + \lambda l_2 \right)x + \left( m_1 + \lambda m_2 \right)y + \left( n_1 ' + \lambda n_2 \right) = 0\]

\[ \Rightarrow \text { Slope of diagonal AC } = - \left( \frac{l_1 + \lambda l_2}{m_1 + \lambda m_2} \right)\]

Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by
l₁x + m₁y + n₁ + \[\mu\] (l₂x + m₂y + n₂) = 0

\[\Rightarrow \left( l_1 + \mu l_2 \right)x + \left( m_1 + \mu m_2 \right)y + \left( n_1 + \mu n_2 \right) = 0\]

\[ \Rightarrow \text { Slope of diagonal BD  }= - \left( \frac{l_1 + \mu l_2}{m_1 + \mu m_2} \right)\]

The diagonals are perpendicular to each other.
∴ \[\left( \frac{l_1 + \lambda l_2}{m_1 + \lambda m_2} \right)\left( \frac{l_1 + \mu l_2}{m_1 + \mu m_2} \right) = - 1\]

\[\Rightarrow \left( l_1 + \lambda l_2 \right)\left( l_1 + \mu l_2 \right) = - \left( m_1 + \lambda m_2 \right)\left( m_1 + \mu m_2 \right)\]

\[\text { Let  }\lambda = - 1, \mu = 1\]

\[ \Rightarrow \left( l_1 - l_2 \right)\left( l_1 + l_2 \right) = - \left( m_1 - m_2 \right)\left( m_1 + m_2 \right)\]

\[ \Rightarrow \left( {l_1}^2 - {l_2}^2 \right) = - \left( {m_1}^2 - {m_2}^2 \right)\]

\[ \Rightarrow \left( {l_1}^2 - {l_2}^2 \right) + \left( {m_1}^2 - {m_2}^2 \right) = 0\]